Solve (sqrt{3})^2=2^2+1^2-2*2*x

$$( \sqrt { 3 } ) ^ { 2 } = 2 ^ { 2 } + 1 ^ { 2 } - 2 \times 2 \times x$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps for Solving Linear Equation

The square of \sqrt{3} is 3.

$$3=2^{2}+1^{2}-2\times 2x$$

Calculate 2 to the power of 2 and get 4.

$$3=4+1^{2}-2\times 2x$$

Calculate 1 to the power of 2 and get 1.

$$3=4+1-2\times 2x$$

Add 4 and 1 to get 5.

$$3=5-2\times 2x$$

Multiply 2 and 2 to get 4.

$$3=5-4x$$

Swap sides so that all variable terms are on the left hand side.

$$5-4x=3$$

Subtract 5 from both sides.

$$-4x=3-5$$

Subtract 5 from 3 to get -2.

$$-4x=-2$$

Divide both sides by -4.

$$x=\frac{-2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out -2.

$$x=\frac{1}{2}$$